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An interesting trigonometric integral

Once while solving problems from Romanian Mathematical Olympiads I encountered a mathematical gem. The problem I found was asking to calcualte the following integral $I=\int\frac{\cos x}{a\sin x+b\cos x}\,dx.$

Solution. Let's try to evaluate one more integral $J=\int{\frac{\sin x}{a\sin x+b\cos x}}\,dx.$

Can you guess the next step?

Yes, we will take advantage of the fact that the sum of integrals is the integral of sum. In other words: $aJ+bI=a\int{\frac{\sin x}{a\sin x+b\cos x}}\,dx+b\int{\frac{\cos x}{a\sin x+b\cos x}}\,dx\\=\int{\frac{a\sin x+b\cos x}{a\sin x+b\cos x}}\,dx=x+C.$

On other hand we can easily calculate integrals of the form $$\int{\frac{f^\prime(x)}{f(x)}}\,dx$$. The derivative of $$(a\sin x+b\cos x)$$ is equal to $$(a\cos x-b\sin x)$$, which can also be expressed as a linear combination of our two integrals, i.e. $aI-bJ=\int{\frac{a\cos x-b\sin x}{a\sin x+b\cos x}}\,dx=\ln{|a\sin x+b\cos x|}+C.$

Now we simply have to solve a system of linear equations. We obtain $I=\frac{a\ln{|a\sin x+b\cos x|}+bx}{a^2+b^2}+C.$ and $J=\frac{ax-\ln{|a\sin x+b\cos x|}}{a^2+b^2}+C.$

Has anyone of you seen problems with the similar ideas? How often while solving a problem you investigate possible variations of the conditions?

Note by Nicolae Sapoval
2 years, 10 months ago

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Try solving this the same way: $$\Large I = \int \frac {a \sin x + b \cos x }{c \sin x + d \cos x} \,dx$$ · 2 years, 10 months ago

1. Lets split the integral $$I$$ into the sum of two integrals: $I=a\int\frac{\sin x}{c\sin x+d\cos x}\,dx+b\int\frac{\cos x}{c\sin x+d\cos x}\,dx=aJ+bK.$
2. Now we reduced our problem to the previous one, because integrals $$J,K$$ are exactly the same as the one mentioned in the post.
· 2 years, 10 months ago

Smiley_Face.gif · 2 years, 10 months ago

This looks very useful, thanks for sharing Nicolae. :) · 2 years, 10 months ago

Thanks for sharing !! · 2 years, 10 months ago